\(\int \frac {\csc ^3(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a-3 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{5/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}} \] Output:

-1/2*(a-3*b)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(5/2 
)/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(1/2)-3/2*b*sec(f*x 
+e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(304\) vs. \(2(127)=254\).

Time = 2.74 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.39 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\frac {(a+3 b+(a-3 b) \cos (2 (e+f x))) \csc ^2(e+f x) \sec (e+f x)}{\sqrt {2} a^2 \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}+\frac {(a-3 b) \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 a^{5/2} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}}{2 f} \] Input:

Integrate[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(-(((a + 3*b + (a - 3*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2*Sec[e + f*x])/(S 
qrt[2]*a^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])) + ((a 
 - 3*b)*Cos[e + f*x]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x 
)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e 
+ f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/ 
2]^2)^2]])*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[ 
e + f*x]^2])/(2*a^(5/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f 
*x)/2]^4]))/(2*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 373, 402, 25, 27, 291, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \tan (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4147

\(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int \frac {-2 b \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{2 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\frac {3 b \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {(a-3 b) (a-b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}}{2 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\frac {\int \frac {(a-3 b) (a-b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}+\frac {3 b \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{2 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\frac {(a-3 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {3 b \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{2 a}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\frac {(a-3 b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {3 b \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{2 a}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\frac {(a-3 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {3 b \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{2 a}}{f}\)

Input:

Int[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(Sec[e + f*x]/(2*a*(1 - Sec[e + f*x]^2)*Sqrt[a - b + b*Sec[e + f*x]^2]) - 
(((a - 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]] 
)/a^(3/2) + (3*b*Sec[e + f*x])/(a*Sqrt[a - b + b*Sec[e + f*x]^2]))/(2*a))/ 
f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(933\) vs. \(2(111)=222\).

Time = 6.21 (sec) , antiderivative size = 934, normalized size of antiderivative = 7.35

method	result	size
default	\(\text {Expression too large to display}\)	\(934\)

Input:

int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/f/a^(7/2)*(((1-cos(f*x+e))^6*csc(f*x+e)^6-(1-cos(f*x+e))^4*csc(f*x+e)^ 
4-(1-cos(f*x+e))^2*csc(f*x+e)^2+1)*a^(5/2)+(12*(1-cos(f*x+e))^4*csc(f*x+e) 
^4+12*(1-cos(f*x+e))^2*csc(f*x+e)^2)*b*a^(3/2)+4*ln(2/(1-cos(f*x+e))^2*(-a 
*(1-cos(f*x+e))^2+2*(1-cos(f*x+e))^2*b+2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/ 
(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2))*((a*cos(f*x+ 
e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^2*(1-cos(f*x+e))^2*csc(f*x+ 
e)^2-12*ln(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2+2*(1-cos(f*x+e))^2*b+2* 
((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e 
)^2+a*sin(f*x+e)^2))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1 
/2)*a*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+4*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+ 
e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b 
*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b 
)/(cos(f*x+e)+1))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2) 
*a^2*(1-cos(f*x+e))^2*csc(f*x+e)^2-12*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e) 
^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*s 
in(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/ 
(cos(f*x+e)+1))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a 
*b*(1-cos(f*x+e))^2*csc(f*x+e)^2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1- 
cos(f*x+e))^2*csc(f*x+e)^2+4*(1-cos(f*x+e))^2*b*csc(f*x+e)^2+a)/(1-cos(f*x 
+e))^2*sin(f*x+e)^2/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^3/(a+b*tan(f*x+e)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (111) = 222\).

Time = 0.26 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.72 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, -\frac {{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \] Input:

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/4*(((a^2 - 4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f 
*x + e)^2 - a*b + 3*b^2)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 + 2*sqrt(a 
)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/ 
(cos(f*x + e)^2 - 1)) - 2*((a^2 - 3*a*b)*cos(f*x + e)^3 + 3*a*b*cos(f*x + 
e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 - a^3*b)*f*co 
s(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*x + e)^2), -1/2*(((a^2 - 
4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 - a*b 
 + 3*b^2)*sqrt(-a)*arctan(-sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos( 
f*x + e)^2)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b)) - ((a^2 - 3*a*b)*co 
s(f*x + e)^3 + 3*a*b*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f 
*x + e)^2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f* 
cos(f*x + e)^2)]
 

Sympy [F]

\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(csc(f*x+e)**3/(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Integral(csc(e + f*x)**3/(a + b*tan(e + f*x)**2)**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 731 vs. \(2 (111) = 222\).

Time = 1.71 (sec) , antiderivative size = 731, normalized size of antiderivative = 5.76 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

1/8*((((a^5*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^4*b^2*sgn(tan(1/2*f*x + 
1/2*e)^2 - 1))*tan(1/2*f*x + 1/2*e)^2/(a^6*b - a^5*b^2) - 2*(a^5*b*sgn(tan 
(1/2*f*x + 1/2*e)^2 - 1) - 7*a^4*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 6*a 
^3*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^6*b - a^5*b^2))*tan(1/2*f*x + 1 
/2*e)^2 + (a^5*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 7*a^4*b^2*sgn(tan(1/2*f 
*x + 1/2*e)^2 - 1) - 8*a^3*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^6*b - a 
^5*b^2))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b* 
tan(1/2*f*x + 1/2*e)^2 + a) - 4*(a - 3*b)*arctan(-(sqrt(a)*tan(1/2*f*x + 1 
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b 
*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*f*x + 1/ 
2*e)^2 - 1)) - 2*(a - 3*b)*log(abs(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt 
(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 
 1/2*e)^2 + a))*sqrt(a) + a - 2*b))/(a^(5/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 
1)) + 2*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 
 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a - 2*(sqrt 
(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f 
*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b - a^(3/2))/(((sqrt(a)*t 
an(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 
1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)*a^2*sgn(tan(1/2*f*x + 1 
/2*e)^2 - 1)))/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(3/2)),x)
 

Output:

int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{4} b^{2}+2 \tan \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x)**3)/(tan(e + f*x)**4*b**2 + 
2*tan(e + f*x)**2*a*b + a**2),x)